JEE Main Chemistry Question (2026) — Solution

Step-by-step solution

To determine which compounds are soluble in aqueous NaOH, we need to identify the compounds that are sufficiently acidic to react with NaOH (a strong base) to form water-soluble sodium salts. Generally, carboxylic acids and phenols are acidic enough to dissolve in aqueous NaOH, whereas alcohols, aldehydes, amines, and hydrocarbons are not. Let us analyze each compound: (I) Benzaldehyde: It is an aldehyde. It does not have an acidic proton and is insoluble in aqueous NaOH. (II) 1-Naphthol: It is a phenol derivative. Phenols are acidic and react with NaOH to form soluble phenoxide salts. (Soluble) (III) 4-(Dimethylamino)phenol: It contains a phenolic -OH group. Despite the basic amino group, the acidic phenol group will react with NaOH to form a soluble salt. (Soluble) (IV) p-Cresol: It is a phenol derivative. It reacts with NaOH to form a soluble sodium salt. (Soluble) (V) Benzoic acid: It is a carboxylic acid, which is highly acidic and readily dissolves in aqueous NaOH. (Soluble) (VI) N,N-Dimethylcyclohexanamine: It is an aliphatic amine. It is basic in nature and insoluble in aqueous NaOH. (VII) 1-Naphthoic acid: It is a carboxylic acid and will dissolve in aqueous NaOH. (Soluble) (VIII) 1,4-Di-tert-butylbenzene: It is a hydrocarbon. It is neutral and insoluble in aqueous NaOH. (IX) 1-Naphthalenemethanol: It is an alcohol (benzyl-type). Alcohols are generally not acidic enough to react with aqueous NaOH. (Insoluble) The compounds soluble in aqueous NaOH are (II), (III), (IV), (V), and (VII). Total number of soluble compounds = 5. Answer: 5